Prove that a transformation from the hyperbolic group can not be loxodromic.

Question

Prove that a transformation from the hyperbolic group can not be loxodromic. I know a loxodromic λ = kei$^\theta$ with k not equal to 1 and theta not equal to 0. But I'm unsure how to go after that, I actually thought you could have a loxodromic and so I'm confused. Is it that loxodromic transformations are homothetic and that the hyperbolic group has to remain inside the unit disk and so a loxodromic transformation would alter the unit disk? I'm quite confused, if anyone could offer what the first step to take might be I'd really appreciate it.

Answer 1

A linear fractional trnasformation that carries the unit disc to itself, and boundary to boundary, is of general form $$ f(z) = \left( \frac{\alpha z + \beta}{\bar{\beta} z + \bar{\alpha} } \right), $$ where the requirement of taking interior to interior is that $$ | \beta| < |\alpha|, $$ so that $|f(0)| < 1.$

If $\alpha \neq 0$ we can divide through by it to get $$ f(z) = e^{i \theta} \; \left( \frac{ z + \gamma}{\bar{\gamma} z + 1} \right) $$ with $ | \gamma| < 1, $ or when $\beta \neq 0$ we can divide by that and get $$ f(z) = e^{i \tau} \; \left( \frac{ \delta z + 1}{z + \bar{\delta} } \right) $$ with $ | \delta| > 1. $

These are all exactly the same, just a matter of convenience. If you can tell me, in one of three common recipes, which is "loxodromic," I can tell you what happens next.

Oh, if preferred, we may divide through instead by either the real valued $|\alpha|$ or the real valued $|\beta|,$ in which case we do not need any factor outside the parentheses.

Answer 2

Since you already know that a loxodromic transformation is conjugate to multiplication by a non-unimodular, non-positive number $ke^{i\theta}$, a logical approach is to show that such a transformation cannot preserve a disk on the Riemann sphere. That is, there is no disk/halfplane $D$ such that $\{ke^{i\theta}z:z\in D\}=D$. Since $k\ne 1$, it follows that if $D$ exists, it must be a halfplane with $0$ on the boundary. But since $\theta\ne 0 \mod 2\pi$, a halfplane does not work either.